If z and omega are two complex numbers such that

Given $|z \omega|=1$.............(1)

and $\arg \left(\frac{z}{\omega}\right)=\frac{\pi}{2}$.............(2)

$\therefore \frac{z}{\omega}+\frac{\bar{z}}{\omega}=0$ $\left[\because \operatorname{Re}\left(\frac{z}{\omega}\right)=0\right]$

$\Rightarrow z \bar{\omega}=-\bar{z} \bar{\omega}$

from equation (i), $z \bar{z} \omega \bar{\omega}=1$ [using $z \bar{z}=|z|^{2}$ ]

$(\bar{z} \omega)^{2}=-1 \Rightarrow \bar{z} \omega=\pm i$

from equation (ii), $-\arg (\bar{z})-\arg \omega=\frac{\pi}{2}-\arg (\bar{z} \omega)=\frac{-\pi}{2}$

Hence, $\bar{z} w=-i$